
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 6:28 PM


On Dec 3, 4:43 pm, jaakov <removeit_jaakov@deleteit_ro.ru> wrote: > On 03.12.2012 20:08, Butch Malahide wrote: > > On Dec 3, 8:21 am, jaakov<removeit_jaakov@deleteit_ro.ru> wrote: > > >> Given a set X and a cardinal k, is there a set Y such that card(Y)=k and > >> X is disjoint from Y? > > > This is equivalent to asking, for a given set X, is there a set Y such > > that card(Y) = card(X) and X is disjoint from Y? > > > Namely, given a set X and a cardinal k, we can take a set K with > > card(K) = k and let X' be the union of X and K. If we can find a set > > Y' such that card(Y') = card(X') and Y' is disjoint from X', then Y' > > has a subset Y such that card(Y) = k, and of course Y is disjoint from > > X. > > >> Is there a proof of this fact that works without the axiom of regularity > >> (= axiom of foundation) and does not assume purity of sets? > > > Let X be a given set. For each set S in P(X), let Y_S = {(S,x): x in > > X}. Clearly Y_S = X. Assuming that X meets Y_S for each S in P(X), > > we could define a surjection from X to P(X), > > How?
Consider any x in X. If x is an ordered pair (u,v) with u in P(X), define f(x) = u; otherwise define f(x) = X. This defines a function f:X > P(X). If Y_S has nonempty intersection with X, then S is in the range of f.
> > A more concrete version of this argument, a la Russell: Given a set X, > > let > > T = {(S,x): S in P(X), x in X, (S,x) in X, (S,x) not in S} > > and let Y = {(T,x): x in X}. Clearly Y = X. Assuming X is not > > disjoint from Y, there is an element x in X such that (T,x) is in X. > > Now we get the Russell paradox in the form > > (T,x) is in T<> (T,x) is not in T. > > Not quite. You have > (T,x) in T <=> T in P(X) and (T,x) not in T. > This is not yet a contradiction.
From the definition of T, namely
T = {(S,x): blah, blah, (S,x) in X, blah}
it can be seen that T is a subset of X, that is, T is an element of P(X). Inasmuch as "T in P(X)" is true, the statement "T in P(X) and (T,x) not in T" simplifies to "(T,x) not in T".
> Thank you anyway.

